CAT 2005 — QA Question 18

${\log }_x\left(\frac{x}{y}\right)+{\log }_y\left(\frac{y}{x}\right)$ = log_x x - log_x y + log_y y - log_y x

⇒ ${\log }_x\left(\frac{x}{y}\right)+{\log }_y\left(\frac{y}{x}\right)$ = 1 - log_x y + 1 - log_y x

⇒ ${\log }_x\left(\frac{x}{y}\right)+{\log }_y\left(\frac{y}{x}\right)$ = 2 - log_x y - log_y x

⇒ ${\log }_x\left(\frac{x}{y}\right)+{\log }_y\left(\frac{y}{x}\right)$ = 2 - (log_x y + log_y x)

As x ≥ y and y > 1,
log_y x ≥ 0
Now, (log_x y + log_y x) = ${\log }_x\left(y\right)+\frac{1}{{\log }_x\left(y\right)}$ [This is sum of a positive number and its reciprocal]

Now, sum of a positive number and its reciprocal is always greater than or equal to 2.

∴ (log_x y + log_y x) = ${\log }_x\left(\frac{x}{y}\right)+{\log }_y\left(\frac{y}{x}\right)$ ≥ 2

⇒ ${\log }_x\left(\frac{x}{y}\right)+{\log }_y\left(\frac{y}{x}\right)$ = 2 - (log_x y + log_y x) ≤ 0

∴ ${\log }_x\left(\frac{x}{y}\right)+{\log }_y\left(\frac{y}{x}\right)$ ≠ 1

Hence, option (d).